Voltage Level Display of 12V Battery

The voltage monitoring of a 12V battery is especially necessary when a discharged battery is being charged. It may be for the batteries used with inverters in home or commercial batteries. Here is a simple project produced to monitor the battery voltage using 741 op amp. The Circuit Diagram is shown in Circuit Diagram Tab.

Generally a 12V battery is expected to produce a 12V dc voltage all the time until it gets discharged. Practically, it is not possible and the output voltage of the battery decreases gradually as load connected to it draws current from it. So it is generally considered that if the output voltage of a 12V battery changes from 12V to 9V approximately, it is constant and after the voltage falls below 9V, it is considered to be discharged and need to be charged again. Any inverter designed to use with a 12V battery should be comfortable to produce a constant AC output voltage during the fall of battery voltage from 12V to 9v approximately. My point to say all these is that to monitor a 12V battery, it is sufficient to monitor the battery voltage from 12V to 9V range and a voltage below 9V should give a warning for LOW BATTERY VOLTAGE. Here for the development of the circuit, the 12V to 9V range is divided into 6 parts as 9V, 9.5V, 10V…11.5V. So 6 op amps are used to compare and display the appropriate level of voltage. The circuit is simple. Comparison is done between the two terminals of each op amp and accordingly output is shown. The series of resistors connected at the inverting terminals of each op amp makes a voltage divider and exact comparing voltage is supplied to each op amp.

For example if the battery voltage is 10.7V, the upper 2 LEDs in the circuit will be OFF and the lower 4 LEDs will be ON. Similarly if the battery voltage is 9.2V only the LED at the bottom will be ON and other 5 will be OFF.

If the voltage goes down below 9V, the transistor will be activated and the buzzer will start indicating LOW VOLTAGE.

The value of the resistors is calculated as

Assume current ‘i’ flowing through the voltage divider is 1 mA. There should be six identical resistors across each op amp. Let the value of each resistor be R.

But we need to monitor voltage from 9V to 12V i.e. only 3V.So the extra 9V should be dropped across somewhere. Let it be resistor X.

Then (6*R + X)*1*10^-3 =12V

Let, R=0.5K

Then, X=9K

The power rating of the resistors should be above 2 Watt.

COMPONENTS

1.Op Amp 741 [6]

2.Resistor (5K6 [6], 0.5 watt)

3.Resistor (0.5K [6], 9K [1], 2 watt) (9K if not available may be split to three 3K )